How is this a telescoping series?

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The series in question is $$ \sum_{k=1}^{\infty} \frac{1}{(2k-1)(2k+1)} $$, which is claimed to be telescoping with a limit of ##\frac{1}{2}##. Initial confusion arose from incorrectly splitting the term into $$ \frac{1}{(2k-1)} + \frac{1}{(2k+1)} $$, leading to a misunderstanding of how telescoping works. Upon further examination, it was clarified that the correct approach involves recognizing the algebraic relationship between the terms rather than relying on intuition. The discussion concluded with an acknowledgment of the algebraic error and a better understanding of the series' behavior.
arhzz
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Homework Statement
Finding the limit of the series
Relevant Equations
Telescoping series formula
Hello !

Consider this series;

$$ \sum_{k=1}^{\infty} \frac{1}{(2k-1)(2k+1)} $$ It is said to find the limit of the series when approaches infinity.Now it is said that this is a telescopic series and that the limit is ##\frac{1}{2}## but I don't see it. I've split the an part (I don't know how to call it in english but it is the term within the sum symbol) I did that like this.

$$ \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$ Now if we input k to let's say 3 we get this

$$ \frac{1}{2} + \frac{1}{3} +\frac{1}{3} + \frac{1}{5} +\frac{1}{7} + \frac{1}{7} $$. I think its fair to assume that only 1/2 should remain and the others should cancel out,that is kind of the thing by the telescoping series.But I have + across the board and the fractions won't cancel out but rather add up.I am not sure how the - is supposed to come into play here,since as I've been taught yestarteday we have to sum up the elements.The only logical solution would be that within the parantheses one element has to be negative,but I don't see how we get to that.
 
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arhzz said:
$$ \sum_{k=1}^{\infty} \frac{1}{(2k-1)(2k+1)} $$ It is said to find the limit of the series when approaches infinity.Now it is said that this is a telescopic series and that the limit is ##\frac{1}{2}## but I don't see it. I've split the an part (I don't know how to call it in english but it is the term within the sum symbol) I did that like this.

$$ \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$
Are you sure about that?
 
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Why do you think

$$ \frac{1}{(2k-1)(2k+1)} = \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$ ?

Plug in a value for k. Like 2.
 
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Vanadium 50 said:
Why do you think

$$ \frac{1}{(2k-1)(2k+1)} = \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$ ?

Plug in a value for k. Like 2.
Okay so you are definately right,these are not the same so the algebra doesn't add up.I supposed it would have to be 1st fraction - 2nd fraction but I assumed it would be + because I don't know intuition? I saw * in the denominator and simply assumed it would be +
 
Don't use intuition. Use algebra.
 
Vanadium 50 said:
Don't use intuition. Use algebra.
Yea I guess I rushed it. It all makes sense now,thanks for pointing out the mistake.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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